$AB_1B_2$ is the base of a pyramid $DAB_1B_2$. $$\pi:2x+y+2z-15=0$$ $$A(3,-2,1)\,\,C(6,-2,-2) \,\,B_1(5,2,-3) \,\,B_2(7,-6,-1)$$ $D$ is on plane $\pi$ and this plane is parallel to plane $$AB_1B_2$$.
First I wanted to find the area of the triangular basis so I started with the height to $B_1B_2$, $\,\,\overrightarrow{AE}$.
I found that the line $B_1B_2$ is:
$$(5,2,-3) + t(2,-8,2)$$
A general point on the line would be $$E(5+2t,2-8t,-3+2t)$$ and AE would be
$$\overrightarrow{AE} = (2+2t,4-8t,-4+2t)$$
AE is a height so $$\overrightarrow{B_1B_2}\cdot\overrightarrow{AE}=(2+2t,4-8t,-4+2t)\cdot(2,-8,2)=0$$ $$72t = 36 \Rightarrow t=2$$
$$\overrightarrow{AE} = (6,-12,0)$$ $$\left\lvert\overrightarrow{AE}\right\rvert = 6\sqrt{5},\,\,\left\lvert\overrightarrow{B_1B_2}\right\rvert = 6\sqrt{2}$$
Because $AB_1B_2$ and $\pi$ are parallel, the distance from each of the points $A$, $B_1$, $B_2$ to the plane $\pi$ will be equal. This distance is calculated and it is $3$.
Then the volume of the the pyramid is $$V = \frac{S_{AB_1B_2}\cdot h}{3}$$
This leads to $V=18\sqrt{10}$ but $V=18$
What did I do wrong?
The issue is that $(\pi)$ is not parallel to $AB_1B_1$ (as claimed in your text) because:
$$\vec{AB_1} \times \vec{AB_2}=\begin{pmatrix}2\\4\\-4\end{pmatrix} \times \begin{pmatrix}4\\-4\\-2\end{pmatrix}=\begin{pmatrix}8\\-12\\-24\end{pmatrix}$$
is not proportional to the normal vector $\begin{pmatrix}2\\1\\2\end{pmatrix}$ of plane $(\pi)$.